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When SbCl3(g) (4.867 mol) and 0.01217 mol/L of Cl2(g) in a 400.0 L reaction vessel at 958.0 K are allowed to come to equilibrium the mixture contains 0.008586 mol/L of SbCl5(g). What concentration (mol/L) of SbCl3(g) reacted?

SbCl3(g)+Cl2(g) = SbCl5(g)

Unit Conversions:
K = C + 273
Molar Mass (g/mol)
SbCl3(g) 228.16
Cl2(g) 70.906
SbCl5(g) 299.06

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